"if the left-hand mass is 2.3 kg ,what should the right-hand mass be so that it accelerates downslope at 0.64 m/s2? "

m₁ = 2.3 kg 
θ₁ = 70° 
θ₂ = 17° 
g = 9.8 m/s² 

->The component of the gravitational force on m₁ that is parallel down the incline is: 
F₁ = m₁ × g × sin(θ₁) 
F₁ = (2.3 kg) × (9.8 m/s²) × sin(70°) = 21.18 N 

->The component of the gravitational force on m₂ that is parallel down the incline is: 
F₂ = m₂ × g × sin(θ₂) 
F₂ = m₂ × (9.8 m/s²) × sin(70°) = m₂ × (2.86 m/s²) 

Then the total mass of the system is: 
m = m₁ + m₂ 
m = (2.3 kg) + m₂ 

If it is given that m₂ slides down the incline, then F₂ must be bigger than F₁, 
and so the net force on the system must be: 
F = m₂×(2.86 m/s²) - (21.18 N) 

Using Newton's second law, we know that 
F = m × a 
So if we want the acceleration to be 0.64 m/s², then 
m₂×(2.86 m/s²) - (21.18 N) = [(2.3 kg) + m₂] × (0.64 m/s²) 
m₂×(2.86 m/s²) - (21.18 N) = (1.47 N) + m₂×(0.64 m/s²) 
m₂×(2.22 m/s²) = (22.65 N) 
m₂ = 10.2 kg

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